w = (1 - iz)/(1 + iz) = (1- ix + y)/(1 + ix - y) == (1-lzl2 + i2x)/(1- 2y + lzi2)
lwi2 = (1 + 2y + lzl2)/(l- 2y + lzl2).
(!1, �' � ) = (-x3, -x 1, x2) so that (!1, �' � ) is obtained as a 90°
counterclockwise rotation about the x2 axis followed by a 90°
counterclockwise rotation about the x-axis.
Any circle or line in the z-plane corresponds to a line or circle on the
stenographic projection onto he Riemann sphere. The function w=l/z
rotates the Riemann sphere 180° about the x 1 axis. Lines and circles on
the rotated sphere project to lines and circles in the w-plane. As a result
lines and circles in the z-plane map to lines and circles in the w-plane.

EXERCISES 2.2: Limits and Continuity

i, - -i, 16, and 32i .
1

1. The first five terms are, respectively,
,
4
sequence converges to O in a spiral-like fashion.

Replace the partial derivatives on the right sides of the equations for
:� and :� by their Cauchy- Riemann counterparts to obtain:
1 ov
ov
ov .
OU
- = cos() - s B = --

or

fJv

-

- m

oy
ox
Bu .
Bu
= - cosB+ s B

-

or

oy

r {)()
1 ou

= --•

- m

r 80

ox

7. Let h(z) = f(z) - g(z). Then his analytic in D and h'(z) = 0 so his
a constant function.

h(z)

= c = f(z) - g(z)

� f(z) = g(z) + c

.
ou
fJu
fJv
8. u(x,y)=cmD� -=Oand-=0=--. Hence
ox
fJy
ox

f'(z) = ::

+

i::

= 0 so

f is constant in D.

9. By contradiction. If f is analytic in a domain D then v( x, y) = 0 (a
constant)=> f is constant (by condition 8) => u is constant. (However,
there is no open set in which u(x, y) = jz2 - z] is constant).
.

13. f/(z)I is analytic and real-valued, so the result follows from Exercises
10 and 12.
14. If the line is vertical then Re/(z) is constant a.nd this reduces to Problem 8. H the line is not vertical, then v(x,y) = mu(x,y) + b, and

2 f. It is easily verified that u = In lzl satisfies Laplace's equation on
C\ {O} and that u+iv = ln lzl+iArg(z) satisfies the Cauchy-Riemann
equations on the domain D = C\ { nonpositive real axis}, so that
Arg (z) is a harmonic conjugate of u on D. By Problem 4, any harmonic
conjugate of u has to be of the form Arg( z) + a in D. It is impossible to
have a. harmonic . conjugate of this form that is continuous on C \ { 0}.

EXERCISES 2.6: Steady-State Temperature as a Harmonic Func•
tion.
1.

b.

a..

100°

100°

scr

800

80°

Mr

6()°

60°
40°

40°

40°

2()°

20°

20°

80°
60°
40°
20,

()°
()°
()°

d.

c.
2(1'

40° 60° 8<>°

500 I 00° 50°

100°

goo
60°
40°

oo

2()°

er

er

e.
100°
..._ 80°

80°
60°........__

._ .

60°

40°

40°

20°

20"

oo

oo

2. This does not violate the maximum principle.

oo

3. This does not violate the maximum principle.
100Â°

100Â°

Exercises 2. 7
1.

2.

3.

f(z) = z2 + c where c is a real constant.
St= (I + "(1-4c))/2, s2 = (1 - "(1-4c))/2
Only s2 is an attractor for -3/4 < c < 1.4.
f(<;) =<;and f'(<;) > 1 Therefore we can pick a real number p between 1
and If(<;) I such that lf(z) - <;I= plz-<;I for all z in a sufficiently small disk
around q, If any point z0 in this disk is the seed for an orbit z1 = f(zo), z2 =
F(z.), ... Zn= f(Zn-t), then we have lzn-<;1 � plzn-t-<;I � ... � pnlZn-t-<;I.
Because p>l, the point Zn moves away from <; until the
magnitude of the derivative becomes 1 or less. The orbit is out of the disk.
(a) Fixed points are SI = i, s2 = -i. Both are repellors.
(b) Fixed points are St= 1/2, s2 = -1/2, s = -1. Fixed points
and SJ
are repellors, but fixed point s2 is an attractor.
zo = ei2rra with a an irrational real number. Zn = ei2rra 2"n. Because lznl = 1,
the trajectory will follow the unit circle. If iterations p and q coincide,
21ta2P - 21ta2q = 21ta(2P - 2q) = 21tk for some integer k. But because (2P 2q) is an integer that can be represented by m, the equation 21tam=21tk is
satisfied only if k=am or a= k/m. Because a is irrational it cannot be
represented by a rational number and no iterations repeat.
Fixed points are SI= -1/2 + i"5/2 (an attractor) and s2 = -1/2 - i"5/2 (a
repellor).
f(z) = z2. The seed is zo. ZI = zl, z2 = z04, ... Zn= zoA(2n). To have an n
cycle Zn= zo = zoA(2n). Or Znlzo = zoA(2n -1) = 1 = ei2rr. Solving gives
Zo = eA(i27t/(2n- l )).
The cycle is 4. 2\27tlp) = 21t mod p => 24 = 1 mod p. p=3,5, 15. 3 will
give repeated cycles of length 2. 5 and 15 will give the desired cycles of
length 4.
Student Matlab: n=lOO;c=.253; zo=O;y(l)=zo;
for k= 1 :n-1,y(k+ 1 )=y(k)A2+c;end
plot(y)
If lal s; 1 the whole complex plane is the filled Julia set. If lal 2:: I the
origin is the filled Julia set.
f(z) = z - F(z)/F'(z). f(s) = s - F(s)/F'(s) = s=> F(s)/F'(s) = 0 =>F(s)=O
with the possible exception of the points where F'(s) = 0.
f'(z) = 1 - F'(z)/F'(z) + F(z)F"(z)/(F'(z))2 = F(z)F"(z)/(F'(z))2
f(s) = F(s)F"(s)/(F'(s))2 = 0 where F'(s) #0 and every zero of F(z) is an
attractor as long as F'(s) # 0.